Timeline for Can an ideal in the ring of holomorphic functions on the complex plane be non-finitely generated?
Current License: CC BY-SA 4.0
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| Oct 30 at 0:54 | comment | added | Jesse Elliott | It might be useful to observe that $(R)$ is a Bezout domain, that is, it is an integral domain in which every finitely generated ideal is principal. So, to show that an ideal is not finitely generated, it suffices to show that it is not principal. See the "principal ideal theorem" in Section 4 of diva-portal.org/smash/get/diva2:305238/FULLTEXT01.pdf (If every ideal of $(R)$ were principal, then $(R)$ would be a PID!) | |
| Oct 29 at 23:11 | comment | added | Tom Goodwillie | This ring is not noetherian. As in David Speyer's comment, there is a sequence of ring elements $f_1,f_2,\dots$ such that $f_n$ is a multiple of $f_{n+1}$ but not conversely. The ideals $(f_n)$ form a strictly increasing sequence whose union cannot be finitely generated. Another example would be $sin\frac{z}{2^n}$. | |
| Oct 29 at 20:58 | comment | added | David E Speyer | Concretely, the ideal that Zerox gives is generated by $\{ \sin \pi z, \tfrac{\sin \pi z}{z}, \tfrac{\sin \pi z}{(z-1)z(z+1)}, \tfrac{\sin \pi z}{(z-2)(z-1)z(z+1)(z+2)}, \cdots \}$. Each of these functions individually has a Weierstrass factorization, but that doesn't address whether the ideal is finitely generated. | |
| Oct 29 at 20:00 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
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| Oct 29 at 19:04 | review | Close votes | |||
| Nov 3 at 3:05 | |||||
| Oct 29 at 18:53 | comment | added | Zerox | Your ideal is the zero ideal since no non-constant holomorphic function can be bounded on $\mathbb{C}$ by Liouville theorem. | |
| Oct 29 at 18:50 | comment | added | Zerox | In my first example, any function in the ideal generated by a finite set of holomorphic functions that vanish on all but finitely many points in $\mathbb{Z}$ can only obtain nonzero values on the union $S$ (which is finite) of the nonzero-valued points of the generators. So this finite generated ideal will not contain the holomorphic functions vanishing on $S$ but nonzero elsewhere. | |
| Oct 29 at 18:48 | comment | added | Haze | thanks for your insight. appreciate it! | |
| Oct 29 at 18:46 | comment | added | Haze | Let the ideal ( I ) consist of all holomorphic functions on the complex plane that satisfy the following increasing constraint:I = \{ f \in R \mid |f(z)| < \frac{1}{n}, \quad \forall z \in D_n \text{ and } n \in \mathbb{Z}_{>0} \},where ( {D_n}_{n=1}^{\infty} ) is an increasing sequence of regions in the complex plane (for example, ( D_n ) could be the disk centered at the origin with radius ( n )). In other words, each holomorphic function ( f ) in ( I ) must satisfy a condition approaching zero on the increasingly large regions ( D_n ). can this ideal be non-finitely generated? | |
| Oct 29 at 18:44 | comment | added | Zerox | Weierstrass factorization theorem has nothing to do with ideal generation because infinite product is not allowed in generating (the reason is that algebraically infinite product is not well-difined). | |
| Oct 29 at 18:39 | comment | added | Haze | @Zerox hi,Zerox,thanks for your comment. I used to think the same; however, I recently came across the Weierstrass factorization theorem, which allows us to construct holomorphic functions that vanish on a given set of zeros, such as the set of positive integers. It seems that this could allow for the construction of a finitely generated ideal ( I ) under such conditions. Do you have any insights on this, or could you provide a reference supporting your conclusion? | |
| Oct 29 at 18:39 | comment | added | Zerox | The ideal of the holomorphic functions that vanish on all but finitely many points in $\mathbb{Z}$ can not be finitely generated. I think every non-principal ultrafilter $\mathcal{F}$ on $\mathbb{Z}$ also gives such an example (the ideal is given by the holomorphic functions that vanish on some set in $\mathcal{F}$). | |
| Oct 29 at 18:25 | history | edited | YCor |
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| S Oct 29 at 18:23 | review | First questions | |||
| Oct 29 at 21:19 | |||||
| S Oct 29 at 18:23 | history | asked | Haze | CC BY-SA 4.0 |